evaluate the definite integral below. ∫₄⁶ (5 + 7ln(x))/x dx

Explanation:

Step1: Use substitution

Let $u = \ln(x)$, then $du=\frac{1}{x}dx$. When $x = 4$, $u=\ln(4)$; when $x = 6$, $u=\ln(6)$. The integral becomes $\int_{\ln(4)}^{\ln(6)}(5 + 7u)du$.

Step2: Integrate term - by - term

$\int_{\ln(4)}^{\ln(6)}(5 + 7u)du=\int_{\ln(4)}^{\ln(6)}5du+\int_{\ln(4)}^{\ln(6)}7udu$. The integral of a constant $C$ is $Cu$, so $\int_{\ln(4)}^{\ln(6)}5du=5u\big|{\ln(4)}^{\ln(6)} = 5(\ln(6)-\ln(4))$. The integral of $au$ (where $a = 7$) is $\frac{au^{2}}{2}$, so $\int{\ln(4)}^{\ln(6)}7udu=\frac{7u^{2}}{2}\big|_{\ln(4)}^{\ln(6)}=\frac{7}{2}(\ln^{2}(6)-\ln^{2}(4))$.

Step3: Combine the results

$5(\ln(6)-\ln(4))+\frac{7}{2}(\ln^{2}(6)-\ln^{2}(4))=5\ln(\frac{6}{4})+\frac{7}{2}(\ln(6) + \ln(4))(\ln(6)-\ln(4))=5\ln(\frac{3}{2})+\frac{7}{2}\ln(24)\ln(\frac{3}{2})$.

Answer:

$5\ln(\frac{3}{2})+\frac{7}{2}\ln(24)\ln(\frac{3}{2})$